1 Answer
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Since I cant comment, I will leave this thought here. Since $||\cdot||_1$ and $||\cdot||$ (as you defined them) are dual norms, it must be that tr$((A+B)^TX)\leq||A+B||_1$ for any $X$ such that $||X||\leq 1$. Therefore, tr$(A^TB)\leq ||A+B||_1 - ||B||^2<||A+B||_1$ (since tr$(B^TB)\geq||B||^2$).

(edit: fixed typo and replaced $||B||$ with $||B||^2$ in the final step)

@Skoro, thanks. Your argument shows that the supremum is at least one. But can it be bigger? For example, are there matrices $A\neq 0$ for which that supremum is at least, say 3? I am interested in bounds relating that quantity to the spectrum of $A$.
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passerby51Jul 8 '13 at 18:53

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Interesting. It looks like one can get an upper bound on $\alpha$ by substituting $B = A/||A||$. This results in the following bound: $\alpha \leq \frac{||A||_1(||A||+1)}{tr(A^TA)} = \frac{(\sum \sigma_i)(\max \sigma_i+1)}{\sum \sigma_i^2}$. maybe this could be controlled using the condition number of $A$.
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SkoroJul 8 '13 at 21:07